Complete the Lewis Dot structure for the molecule H-C≡N and determine how many bonding and non-bonding electrons nitrogen has:

Natural gas burns in air to form carbon dioxide and water, releasing heat. CH4(g)+2O2(g)→CO2(g)+2H2O(g) ΔHrxn = -802.3 kJ.

Olenka [21]

Answer:

1) Minimum mass of methane required to heat 45.0 g of water by 21.0°C is 0.0788 g.

2) Minimum mass of methane required to heat 50.0 g of water by 26.0°C is 0.108 g.

Explanation:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g) ,\Delta H_{rxn} =-802.3 kJ$

1) Minimum mass of methane required to raise the temperature of water by 21.0°C.

Mass of water = m = 45.0 g

Specific heat capacity of water = c = 4.18 J/g°C

Change in temperature of water = ΔT = 21.0°C.

Heat required to raise the temperature of water by 21.0°C = Q

$Q=mc\Delta T= 45.0 g\times 4.18 J/g^oC\times 21.0^oC$

Q = 3,950.1 J = 3.9501 kJ

According to reaction 1 mole of methane on combustion gives 802.3 kJ of heat.

Then 3.950.1 kJ of heat will be given by:

$=\frac{3.950.1 kJ}{802.3 kJ}=0.004923 mol$

Mass of 0.004923 moles of methane :

0.004923 mol × 16 g/mol=0.0788 g

Minimum mass of methane required to heat 45.0 g of water by 21.0°C is 0.0788 g.

2) Minimum mass of methane required to raise the temperature of water by 26.0°C.

Mass of water = m = 50.0 g

Specific heat capacity of water = c = 4.18 J/g°C

Change in temperature of water = ΔT = 26.0°C.

Heat required to raise the temperature of water by 21.0°C = Q

$Q=mc\Delta T= 50.0 g\times 4.18 J/g^oC\times 26.0^oC$

Q = 5,434 J= 5.434 kJ

According to reaction 1 mole of methane on combustion gives 802.3 kJ of heat.

Then 5.434 kJ of heat will be given by:

$=\frac{5.434 kJ}{802.3 kJ}=0.006773 mol$

Mass of 0.006773 moles of methane :

0.006773 mol × 16 g/mol= 0.108 g

Minimum mass of methane required to heat 50.0 g of water by 26.0°C is 0.108 g.

6 0

4 years ago

The half-life for the radioactive decay of C-14 is 5730 years and is independent of the initial concentration. How long does it

DedPeter [7]

Answer:

It will take 2378 years for 25% of the C-14 atoms in a sample of the C-14 to decay.

Explanation:

Radioactive decays/reactions always follow a first order reaction dynamic.

Let the initial amount of C-14 atoms be A₀ and the amount of atoms at any time be A

The general expression for rate of reaction for a first order reaction is

(dA/dt) = -kA (Minus sign because it's a rate of reduction)

k = rate constant

(dA/dt) = -kA

(dA/A) = -kdt

∫ (dA/A) = -k ∫ dt

Solving the two sides as definite integrals by integrating the left hand side from A₀ to A and the right hand side from 0 to t.

We get

In (A/A₀) = -kt

(A/A₀) = e⁻ᵏᵗ

A(t) = A₀ e⁻ᵏᵗ

Although, we can obtain k from the information on half life.

For a first order reaction, the rate constant (k) and the half life (T(1/2)) are related thus

T(1/2) = (In2)/k

T(1/2) = 5730 years

k = (In 2)/5730 = 0.000120968 = 0.000121 /year.

So, the amount of C-14 atoms left at any time is given as

A(t) = A₀ e⁻⁰•⁰⁰⁰¹²¹ᵗ

How long does it take for 25% of the C-14 atoms in a sample of the C-14 to decay?

When 25% of C-14 atoms in a sample decay, 75% of C-14 atoms in the sample remain.

Hence,

A(t) = 75%

A₀ = 100%

100 = 75 e⁻⁰•⁰⁰⁰¹²¹ᵗ

e⁻⁰•⁰⁰⁰¹²¹ᵗ = (75/100) = 0.75

In e⁻⁰•⁰⁰⁰¹²¹ᵗ = In 0.75 = - 0.28768

-0.000121t = -0.28768

t = (0.28768/0.000121) = 2,377.54 = 2378 years

Hope this Helps!!!

3 0

4 years ago

The formal charge is the "charge" an element would have in a molecule or ion if all of the bonding electrons were shared equally

Fiesta28 [93]

Answer:

Explanation:

check below for the solution

5 0

3 years ago

Need help guys!!!!

Serhud [2]

A. When two chemicals mix their temperature rise: Exothermic
b. A solid burns brightly and releases heat, light and sound: Exothermic
c. When two chemicals are mixed their temperature drops: Endothermic
d. Two chemicals will only react if you heat them continually: Endothermic
e. Plants take in light energy for photosynthesis: Endothermic

I hope this helps :)

4 0

4 years ago

"The occurrence of the all-or-none ________________ results from depolarization in response to rapid entry of ____________ ions

Leno4ka [110]

Answer:

The correct answer is

B.action potential; sodium

Explanation:

Action potential

Nerve action potentials are initiated and propagated in an all or nothing fashion. The all-or nothing action potential is initiated when the membrane potential threshold value is reached

Sodium

Voltage-gated Na⁺ ions membrane channels open enabling an influx of sodium ions to rapidly enter the cell, raising the membrane potential up to +40 mV

Depolarization in the neuron is based ion exchange channels opening to allow the rapid influx of sodium ions (Na⁺) and the efflux of potassium ions (K⁺)

The neuronal responds rapidly and for short periods to the activation of an ion channel receptor by a drug or natural neurotransmitter is rapid rel

8 0

3 years ago

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